Question: Let $h(x)=x^{-5}$. $h'(2)=$
Let's first find the expression for $h'(x)$ and then evaluate it at $x=2$. The derivative of $h$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a negative number.) $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-5}\right) \\\\ &=-5x^{-5-1} \gray{\text{The power rule}} \\\\ &=-5x^{-6} \end{aligned}$ So we found that $h'(x)=-5x^{-6}$, which can also be written as $-\dfrac{5}{x^6}$. Now let's plug ${x=2}$ : $\begin{aligned} -\dfrac{5}{({2})^6}&=-\dfrac{5}{64} \end{aligned}$ In conclusion, $h'(2)=-\dfrac{5}{64}$.